What is the general solution for 1+2sin^2(t)=-5cos(t) ?

The general solution for 1+2sin^2(t)=-5cos(t) is t=(2pi)/3+2pin,t=(4pi)/3+2pin

Popular >

1+2sin^2(t)=-5cos(t)

1 + 2 sin^{2} (t) = - 5 cos (t)

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

Solution steps

1 + 2 sin^{2} (t) = - 5 cos (t)

Subtract

- 5 cos (t)

from both sides

1 + 2 sin^{2} (t) + 5 cos (t) = 0

Rewrite using trig identities

3 - 2 cos^{2} (t) + 5 cos (t) = 0

Solve by substitution

cos (t) = - \frac{1}{2}, cos (t) = 3

cos (t) = - \frac{1}{2} : t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

cos (t) = 3 :

No Solution

Combine all the solutions

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

2 sin (2 x) - 3 - \frac{6}{sin ( 2 x ) - 1} = 0

3 sin^{2} (x) + 11 sin (x) + 10 = 0

cos (x) (cos (x) + 5) = 2 sin^{2} (x)

tan^{2} (x) + 3 tan (x) - 10 = 0

cos^{2} (x) - sin (x) = - 1

What is the general solution for 1+2sin^2(t)=-5cos(t) ?
The general solution for 1+2sin^2(t)=-5cos(t) is t=(2pi)/3+2pin,t=(4pi)/3+2pin